Copper wire figures

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AWG Table

1 AWG is 289.3 thousandths of an inch
2 AWG is 257.6 thousandths of an inch
5 AWG is 181.9 thousandths of an inch
10 AWG is 101.9 thousandths of an inch
20 AWG is 32.0 thousandths of an inch
30 AWG is 10.0 thousandths of an inch
40 AWG is 3.1 thousandths of an inch

The table in ARRL handbook warns that the figures are approximate
and may vary dependent on the manufacturing tolerances.
If you don't have a chart handy, you don't really need a formula.
There's several handy tricks:

With these, you can get around alot of different AWGs
and they cross check against one another.
Start with solid 50 AWG having a 1 mil diameter.

So, 30 AWG should have a diameter of ~ 10 mils. Right on with my chart.
36 AWG should have a diameter of ~ 5 mils. Right on with my chart.
24 AWG should have a diameter of ~ 20 mils. Actually ~ 20.1
16 AWG should have a diameter of ~ 50 mils. Actually ~ 50.8
10 AWG should have a diameter of ~ 100 mils. Actually ~ 101.9

If you are more interested in current carrying ability than physical
size, then also remember that a change of 3 AWG numbers equals a
doubling or halving of the circular mills (the cross sectional area).
Thus, if 10 AWG is safe for 30 amps, then 13 AWG (yeah, hard to find) is
ok for 15 amps and 16 AWG is good for 7.5 amps.

The wire gauge is a logarithmic scale base on the cross
sectional area of the wire. Each 3-gauge step in size
corresponds to a doubling or halving of the cross sectional
area. For example, going from 20 gauge to 17 gauge doubles the
cross sectional area (which, by the way, halves the DC
resistance).

So, one simple result of this is that if you take two strands
the same gauge, it's the equivalent of a single wire that's 3
gauges lower. So two 20 gauge strands is equivaent to 1 17
gauge.

Wire Gauge Resistance per foot

Current ratings

Most current ratings for wires (except
magnet wires) are based on permissible voltage drop, not temperature
rise. For example, 0.5 mm^2 wire is rated at 3A in some applications but
will carry over 8 A in free air without overheating.
You will find tables of permitted maximum current in national electrical
codes, but these are based on voltage drop (not the heating which is no
problem in the current rating those codes give).

Here is a small current and AWG table
taken from the Amateur Radio Relay Handbook, 1985.

Mils are .001". "open air A" is a continuous rating for
a single conductor with insulation in open air. "cable amp"
is for in multiple conductor cables. Disregard the amperage
ratings for household use.

To calculate voltage drop, plug in the values: V = DIR/1000
Where I is the amperage, R is from the ohms/1000' column
above, and D is the total distance the current travels (don't
forget to add the length of the neutral and hot together - ie:
usually double cable length). Design rules in the CEC call
for a maximum voltage drop of 6% (7V on 120V circuit).

Resistivities at room temp:

This clearly puts silver as the number one conductor and gold has
higher resistance than silver or copper. It's desireable in connectors
because it does not combine well with other materials so remains
relatively pure at the surface. It also has the capability to adhere to
itself (touch pure gold to pure gold and it sticks together) which makes
for very reliable connections.

Thermal conductivity at room temp:

This explains why diamonds are being used for high power substrates now.
That's man-made diamonds. Natural diamonds contain sufficient flaws in
the lattice that the phonons (heat conductors) get scattered and
substantially reduce the ability to transport the heat.

Wire current handling capacity values

Information about 35 mm2 Cu wire

According Ströberg TTT 35mm2 copper wire can take continuous
current of 170A on free air and 200 A on ground. The wire
can handle 5 kA short circuit current for 1s. DC resistance of
the wiure is 0.52mohm/m.

Mains wiring current ratings

In mains wiring there are two considerations, voltage drop and heat buildup.
The smaller the wire is, the higher the resistance is. When
the resistance is higher, the wire heats up more, and there is
more voltage drop in the wiring. The former is why you need
higher-temperature insulation and/or bigger wires for use in
conduit; the latter is why you should use larger wire for long
runs.

Neither effect is very significant over very short distances.
There are some very specific exceptions, where use of smaller
wire is allowed. The obvious one is the line cord on most
lamps. Don't try this unless you're certain that your use fits
one of those exceptions; you can never go wrong by using larger
wire.

This is a table apparently from BS6500 which is reproduced in the
IEE Wiring Regs which describes the maximum fuse sizes for different
conductor sizes:

Car audio cable recommendations

This info in from rec.audio.car FAQ
(orognally from IASCA handbook).
To determine the correct wire size for your application, you should first
determine the maximum current flow through the cable
(looking at the amplifier's fuse is a relatively simple and
conservative way to do this). Then determine the length of the cable
that your will use, and consult the following chart:

Skin effect

Skin effect is an effect that the electricity in high frequencies does
not use the whole condictor area. High frequencies tend to use only the
outer parts of the conductor. The higher the frequency, the less of the
wire diameter is used and higher the losses. Sin effect must be taken
care in high frequency coil designs.

The frequency dependency of the resistance of a cylindrical conductor
can be calculated by the following formula, which is surely valid for
high frequencies and radii of approx. 50 um:
R(f) = R(DC)* (1 + 1/3 * x^4) with x = Radius/2*sqrt(pi*frequency*permeability*conductivity)

The "formula" for skin effect is the same whether the conductor is
rectangular or cyclindrical. That is why the same value of "radius" used in
wire size in a switchmode transformer is used to determine half the thickness
of a flat foil conductor in the case of foil-wound secondaries.

Skin depth is not an absolute, but only
the depth where current through the wire or foil has fallen to a specific
proportion of the current at the surface. In fact, current falls off
exponenially as you move inward fromm the surface. The depth of the "skin" is
also influenced by proximity to nearby conductors (such as in a transformer)
so is itself not absolute. Also the formula has to be modified if you use wire that is
ferromagnetic (iron for example).

In addition to skin effect a lot of engineers doing their own
magnetics design don't consider the 'proximity effect' which 'crowds'
the current to one side of the conductor and increases losses. This
condition is worst in thick multi-layer windings. Fortunately, many of
the new transformer shapes have a long and skinny window - good for low
leakage L and low proximity effect losses.

Wire sizes used in fuses

The Standard Handbook for Electrical Engineers lists the following formula:

33 * (I/A)^2 * S = log( (Tm - Ta) / (234 + Ta) + 1 )

I = current in Amperes
A = area of wire in circ. mils
S = time the current flows in seconds
Tm = melting point, C
Ta = ambient temp, C

The melting point of copper is 1083 C.
See pp. 4-74 .. 4-79 of the 13th edition of the Handbook for more info.

Skin effect

At high frequencies there is one thing to consider on wire resistance
besides the DC resistence: skin effect.

The current intensity falls off exponentially with depth. The depth of
penetration (s=sigma) is the depth at which the current intensity has
fallen to 1/e of its value at the surface, where e equals 2.718.

Where the diameter of the conductor is large compared to the depth of
penetration, the total current is the same as if the surface current
intensity were maintained to a depth of penetration.